Physics

Student A is on a 9m Hugh bridge while Student B is on the ground. Student B throws an apple up to Student B with a velocity of 18m/s. How long does it take for the apple to reach student A's hand? if student A is unable to catch the apple, how long would it take for the apple to come back to the ground?

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  1. sorry I meant 9m high bridge

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    posted by Sandara
  2. a. h = Vo*t + 0.5g*t^2 = 9
    18t - 4.9t^2 = 9
    -4.9t^2 + 18t - 9 = 0
    t=0.597 s. and 3.08 s(Use quad. formula)
    Choose t = 0.597 s since the apple reaches its max ht. of 16.5 m, in 1.84 s.

    V = Vo + g*t = 0 @ max ht.
    18 - 9.8*t = 0
    9.8t = 18
    Tr = 1.84 s. = Rise time.

    b. If student A does not touch the ball it will reach a max ht. of 16.5 m.:

    h = 18*1.84 - 4.9*1.84^2 = 16.5 m, max.

    Tf = Tr = 1.84 s. = Fall time.

    Tr+Tf = 1.84 + 1.84 = 3.68 s. = Time to
    return to gnd.

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    posted by Henry

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